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 Math Help > Inequalities > Inequality Methods > Schur's Inequality

Schur's Inequality

For all non-negative real numbers x, y, z and a positive number t,
xt(x-y)(x-z) + yt(y-z)(y-x) + zt(z-x)(z-y) ≥ 0

### Example using Schur's Inequality

Prove: for nonnegative a, b, c, 9(a3 + b3 + c3) + 27abc ≥ 2(a + b + c)3

There is a fairly standard approach to symmetric polynomial inequalities.  Write [r,s,t] for the sum of the 6 possible permutations xr ys zt + xs yr zt + ...

Also, the six permutations of (a+b+c)3 = a3 + b3 + c3 + 3a2b + 3ab2 + 3a2c + 3ac2 + 3b2c + 3bc2 + 6abc can be written as

1/2 [3,0,0] + 3[2,1,0] + [1,1,1]

Then our inequality

9(a3 + b3 + c3) + 27abc ≥ 2(a + b + c)3

reduces (that is, by considering the sum of the six inequalities for the six permutations of a,b,c) to

9/2 [3,0,0] + 9/2 [1,1,1] ≥ 2( 1/2 [3,0,0] + 3[2,1,0] + [1,1,1])

or

7/2 [3,0,0] + 5/2[1,1,1] ≥ 6[2,1,0]

Now it's clear that [3,0,0] ≥ [1,1,1] (because (a3+b3+c3)/3 ≥ abc by AM≥GM)  so it suffices to establish that

[3,0,0] + [1,1,1] ≥ 2 [2,1,0]

But this is precisely

x(x-y)(x-z) + y(y-z)(y-x) + z(z-x)(z-y) ≥ 0

which is a case of Schur's inequality, easily established by taking WLOG x ≥ y ≥ z.

### Related pages in this website

Muirhead's Inequality is a general method for proving symmetric polynomial inequalities

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